Visual walkthrough — Related rates — setting up and solving

We will answer this one question:

A straight ladder rests against a wall. Someone pulls its foot away from the wall at a steady speed. How fast is the top of the ladder sliding down the wall at each instant?

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Step 1 — Draw the scene and name what moves

WHAT: we sketch the wall (vertical), the ground (horizontal), and the ladder as the slanted line joining them, then attach a letter to each length.

WHY: we label with letters, not numbers, because the whole point is that $x$ and $y$ move. If we wrote "$x=3$" now, we would freeze the picture and lose the motion — exactly the deadly early-substitution error the parent warns about.

PICTURE: the ladder makes a right triangle. Notice the little square in the corner — that is the guarantee the wall and ground meet at a perfect right angle. Watch that we call the horizontal length $x$ and the vertical length $y$.

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Step 2 — See the two motions as arrows

WHAT: we redraw the ladder with an arrow on the foot pointing away from the wall, and an arrow on the top pointing down the wall.

WHY does $\frac{d}{dt}$ enter and not something else? Because the question is literally "how fast" — a speed. Speed is change ÷ time, and the tool that measures instantaneous change per unit time is the derivative with respect to time. No other tool answers "how fast right now."

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Step 3 — Find the equation that chains $x$ and $y$ together

WHAT: we write down the one equation that ties the two moving lengths to each other.

WHY the Pythagorean theorem and not, say, an angle formula? Because it uses only lengths — and lengths are exactly the quantities whose speeds we want. It also has a fixed right-hand side ($L^2$ never changes), which will be very convenient in Step 5.

PICTURE: the shaded squares built on each side. The blue square on $x$ plus the violet square on $y$ has the same total area as the magenta square on the ladder — at every instant, even as the triangle deforms.

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Step 4 — Realise everything is secretly a function of time

WHAT: we reinterpret $x$ and $y$ as functions of $t$, so that $x^2+y^2=L^2$ holds for all times, not just one photo.

WHY: because a rate ($\frac{dx}{dt}$) only makes sense for something that changes with $t$. To differentiate in time, the thing must first depend on time. This is the same viewpoint as Implicit differentiation: we never solve for $y$ in terms of $t$ — we just accept it is a function of $t$ and differentiate the whole relation.

PICTURE: three stacked freeze-frames — the ladder tipping over as $t$ increases — with $x$ growing and $y$ shrinking.

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Step 5 — Differentiate the link with respect to time

Now apply $\frac{d}{dt}$ to both sides of $x^2+y^2=L^2$.

The Chain rule is the star. To differentiate $x^2$ where $x$ itself depends on $t$, we go through $x$: $\frac{d}{dt}\left(x^2\right)=\underset{\text{outer: }\frac{d}{dx}(x^2)}{\underbrace{2x}}\cdot \underset{\text{inner: rate of }x}{\underbrace{\frac{dx}{dt}}}.$

Doing this to every term:

WHAT: we turned an equation between lengths into an equation between rates.

WHY the $0$ on the right? $L$ never changes, so $L^2$ is a fixed number, and the rate of change of any fixed number is $0$. The ladder's constancy is what makes the two rates add to zero — they must exactly cancel.

PICTURE: the equation drawn as a balance beam. The $x$-term pushes one way, the $y$-term the other; they must sum to nothing, so growth in $x$ forces shrinkage in $y$.

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Step 6 — Solve for the unknown rate

We have $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$ and we want $\frac{dy}{dt}$ alone on one side.

Carrying out the three moves: $y\frac{dy}{dt}=-x\frac{dx}{dt}\stackrel{÷y,y\ne 0}{⟹}\boxed{\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}}$

Plug in one instant. Say $L=5$ m, the foot moves at $\frac{dx}{dt}=1$ m/s, and right now $x=3$ m. From the link, $y=\sqrt{L^2-x^2}=\sqrt{25-9}=4$ m (safely nonzero, so dividing by $y$ was fine). Then $\frac{dy}{dt}=-\frac{3}{4}(1)=-0.75\text{m/s}.$

WHY substitute only now? If we had set $x=3$ back in Step 3, then $x$ would be a constant, $\frac{dx}{dt}$ would be $0$, and our formula would collapse. Numbers go in last — always.

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Step 7 — The degenerate cases (never leave the reader stranded)

The formula $\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$ has a landmine: what if $y=0$? That is exactly the case Step 6 forbade dividing by. Let's walk every extreme.

Let's verify Case B numerically. With $L=5$, $\frac{dx}{dt}=1$ (staying in the first quadrant, $y>0$):

$x$ (m)	$y=\sqrt{25-x^2}$	$\frac{dy}{dt}=-x/y$ (m/s)
$3$	$4.00$	$-0.75$
$4$	$3.00$	$-1.33$
$4.9$	$0.995$	$-4.93$
$4.99$	$0.316$	$-15.8$

The speed blows up — the forecast is confirmed.

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The one-picture summary

One diagram compresses all seven steps: the labelled triangle, the two velocity arrows, the Pythagorean squares, and the final formula with its sign, all in one frame.

Recall Feynman retelling — the whole walkthrough in plain words

Lean a ladder on a wall. Put the corner at the origin, with rightward as positive $x$ and upward as positive $y$; both stay in the first quadrant. Call the floor-gap $x$ and the wall-height $y$; the ladder length never changes. Draw arrows: the foot creeps out to the right (that's a positive speed, $\dot{x}>0$), so the top must creep down (a negative speed, $\dot{y}<0$). Now the magic sentence: the wall, floor and ladder always make a right triangle, so $x^2+y^2$ always equals the ladder squared — a number that never moves. Because $x$ and $y$ are really movies of time, ask "how fast is each side of that equation changing?" On the left, each squared length changes at rate (twice-the-length) × (its own speed) — that's the chain rule paying the "per-second" toll. On the right, a constant changes at zero. So $2x\dot{x}+2y\dot{y}=0$: the two rates must cancel. Divide by two, move the $x$-term over, and divide by $y$ (allowed only while $y$ isn't zero) to get: the top's speed is $-\frac{x}{y}$ times the foot's speed. The minus says "downward"; the fraction $\frac{x}{y}$ says "long base, short height ⇒ terrifyingly fast." And when the ladder lies almost flat, $y$ is tiny, the fraction explodes, and the top rockets down — the equation warns you before the real ladder ever could. The one rule to remember: keep the letters alive while you differentiate, and only pour in the numbers at the very end.

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Connections

• Related rates — setting up and solving — the parent recipe this page derives in pictures.
• Chain rule — the engine that pays the "per-second" toll in Step 5.
• Implicit differentiation — the viewpoint that lets us differentiate $x^2+y^2=L^2$ without solving for $y$.
• Derivatives as rates of change — why $\frac{dx}{dt}$ is a velocity arrow.
• Pythagorean theorem — the linking equation of Step 3.
• Similar triangles — the analogous elimination trick for cone/shadow problems.
• Optimization — the next derivative application after related rates.